the longest side of a right-angled triangle is 13 cm. if the perimeter of the triangle is 30 cm, find the shortest side

The shortest side = 5

Step-by-step explanation:

Call one  side = a

The middle side = b

the hypotenuse = c

a^2 + b^2 = c^2

a^2 + b^2 = 13^2

a^2 + b^2 = 169

a + b + c = 30

a + b + 13 = 30

a + b = 17

b = 17 - a

a^2 + b^2 = 169

a^2 + (17 -  a)^2 = 169

a^2 + (289 - 34a + a^2) = 169

a^2 + a^2 - 34a + 289 = 169           Subtract 169 from both sides

2a^2 - 34a + 289 - 169 = 0

2a^2 - 34a + 120 = 0                      Divide by 2

a^2 - 17a + 60 = 0

This quadratic is easily factored. You need two numbers that add to - 17 and multiply to 60

(a - 5) * (a - 12) = 0    Both give you positive answers. You only want the shortest side

a - 5 = 0

a = 5                                              Is 5 the shortest side?

a - 12 = 0

a = 12                                           Yes 5 is the shortest side.

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.

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step-by-step explanation: